Solve -5x^2+2x+25-9=0

$$- 5 x ^ { 2 } + 2 x + 25 - 9 = 0$$

Solve for x

$$x = -\frac{8}{5} = -1\frac{3}{5} = -1.6$$

$$x=2$$

Steps Using Factoring By Grouping

Subtract 9 from 25 to get 16.

$$-5x^{2}+2x+16=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -5x^{2}+ax+bx+16. To find a and b, set up a system to be solved.

$$a+b=2 ab=-5\times 16=-80$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -80.

$$-1,80 -2,40 -4,20 -5,16 -8,10$$

Calculate the sum for each pair.

$$-1+80=79 -2+40=38 -4+20=16 -5+16=11 -8+10=2$$

The solution is the pair that gives sum 2.

$$a=10 b=-8$$

Rewrite -5x^{2}+2x+16 as \left(-5x^{2}+10x\right)+\left(-8x+16\right).

$$\left(-5x^{2}+10x\right)+\left(-8x+16\right)$$

Factor out 5x in the first and 8 in the second group.

$$5x\left(-x+2\right)+8\left(-x+2\right)$$

Factor out common term -x+2 by using distributive property.

$$\left(-x+2\right)\left(5x+8\right)$$

To find equation solutions, solve -x+2=0 and 5x+8=0.

$$x=2 x=-\frac{8}{5}$$

Steps Using the Quadratic Formula

Subtract 9 from 25 to get 16.

$$-5x^{2}+2x+16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 2 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-5\right)\times 16}}{2\left(-5\right)}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-5\right)\times 16}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-2±\sqrt{4+20\times 16}}{2\left(-5\right)}$$

Multiply 20 times 16.

$$x=\frac{-2±\sqrt{4+320}}{2\left(-5\right)}$$

Add 4 to 320.

$$x=\frac{-2±\sqrt{324}}{2\left(-5\right)}$$

Take the square root of 324.

$$x=\frac{-2±18}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-2±18}{-10}$$

Now solve the equation x=\frac{-2±18}{-10} when ± is plus. Add -2 to 18.

$$x=\frac{16}{-10}$$

Reduce the fraction \frac{16}{-10} to lowest terms by extracting and canceling out 2.

$$x=-\frac{8}{5}$$

Now solve the equation x=\frac{-2±18}{-10} when ± is minus. Subtract 18 from -2.

$$x=-\frac{20}{-10}$$

Divide -20 by -10.

$$x=2$$

The equation is now solved.

$$x=-\frac{8}{5} x=2$$

Steps for Completing the Square

Subtract 9 from 25 to get 16.

$$-5x^{2}+2x+16=0$$

Subtract 16 from both sides. Anything subtracted from zero gives its negation.

$$-5x^{2}+2x=-16$$

Divide both sides by -5.

$$\frac{-5x^{2}+2x}{-5}=-\frac{16}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\frac{2}{-5}x=-\frac{16}{-5}$$

Divide 2 by -5.

$$x^{2}-\frac{2}{5}x=-\frac{16}{-5}$$

Divide -16 by -5.

$$x^{2}-\frac{2}{5}x=\frac{16}{5}$$

Divide -\frac{2}{5}, the coefficient of the x term, by 2 to get -\frac{1}{5}. Then add the square of -\frac{1}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{2}{5}x+\left(-\frac{1}{5}\right)^{2}=\frac{16}{5}+\left(-\frac{1}{5}\right)^{2}$$

Square -\frac{1}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=\frac{16}{5}+\frac{1}{25}$$

Add \frac{16}{5} to \frac{1}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=\frac{81}{25}$$

Factor x^{2}-\frac{2}{5}x+\frac{1}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{5}\right)^{2}=\frac{81}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{5}\right)^{2}}=\sqrt{\frac{81}{25}}$$

Simplify.

$$x-\frac{1}{5}=\frac{9}{5} x-\frac{1}{5}=-\frac{9}{5}$$

Add \frac{1}{5} to both sides of the equation.

$$x=2 x=-\frac{8}{5}$$